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An electron of a hydrogen like atom, having Z = 4, jumps from 4th energy state to 2nd energy state. The energy released in this process, is (Given Rch = 13.6 eV) Where R = Rydberg constant, c = Speed of light in vacuum, h = Planck's constant

1. 3.4 eV
2. 10.5 eV
3. 40.8 eV
4. 13.6 eV

Correct answer: 40.8 eV

Solution

The energy levels of a hydrogen-like atom are given by the formula E_n = -Z² * R_H / n². For Z = 4, the energy difference between the 4th and 2nd states can be calculated, resulting in a release of 40.8 eV when the electron transitions from n=4 to n=2.

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